Bayes Theorem and Ion Saliu’s Paradox

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Erik Bernoth

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Jan 6, 2016, 12:01:35 PM1/6/16

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Bayes Theorem in Poker, Gambling

Since I started playing poker I try to understand, when and why there can be situations, where the probability of an event might not be the same you can get with just counting the probability-tree for each sub-situation. Because I don't understand it, I am not able to apply stuff like the Bayes rule to any problem without someone telling me: "This is a problem for Bayes Theorem".

I show an example:

" Sue and Bob take turns rolling a 6-sided die. Once either person rolls a 6, the game is over. Sue rolls first. If she doesn’t roll a 6, Bob rolls the die; if he doesn’t roll a 6, Sue rolls again. They continue taking turns until one of them rolls a 6.

Bob rolls a 6 before Sue.

What is the probability Bob rolled the 6 on his second turn? "

As I tried to solve this problem, I came to this tree (described as paths, because drawing is not really possible in an email):

Path a) ->Sue1=[6] (1/6); end

Path b) ->Sue1=[1-5] (5/6) -> Bob1=[6](1/6); end

Path c) ->Sue1=[1-5] (5/6) ->Bob1=[1-5] (5/6) ->Sue2=[6] (1/6); end

Path d) ->Sue1=[1-5] (5/6) ->Bob1=[1-5] (5/6) ->Sue2=[1-5] (5/6) -

>Bob2=[6] (1/6); end

Path e) ->Sue1=[1-5] (5/6) ->Bob1=[1-5] (5/6) ->Sue2=[1-5] (5/6) -

>Bob2=[1-5] (5/6); end

Path d is the solution we want. So I count it together and come to:

5/6 * 5/6 * 5/6 * 1/6 = 125/1296 = ~9%

But an experiment with 10^6 games showed ~21% what is 275/1296. So the correct answer must be:

P(A|B) = P(B|A)*P(A)/P(B).

A - Bob wins on his second turn.

B - Bob wins.

Interestingly, it doesn’t matter who goes first. If Sue goes first, as stated, 1*(125/1296)/(5/11) = 275/1296. If Bob goes first, 1*(25/216) / (6/11) = 275/1296.

Questions:

1. Why is the tree-approach not correct here?

2. Where is the logic behind making "bob wins" and "Bob wins second round" 2 different situations that could be put together in the Bayes Theorem?